What is the extraneous solution to these equations? $\dfrac{x^2}{x + 7} = \dfrac{-x + 42}{x + 7}$
Answer: Multiply both sides by $x + 7$ $ \dfrac{x^2}{x + 7} (x + 7) = \dfrac{-x + 42}{x + 7} (x + 7)$ $ x^2 = -x + 42$ Subtract $-x + 42$ from both sides: $ x^2 - (-x + 42) = -x + 42 - (-x + 42)$ $ x^2 + x - 42 = 0$ Factor the expression: $ (x - 6)(x + 7) = 0$ Therefore $x = 6$ or $x = -7$ At $x = -7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -7$, it is an extraneous solution.